ALGEBRAIC EXPRESSIONS AND IDENTITIES 93
8.1 Addition and Subtraction of Algebraic
Expressions
In earlier classes, we have already become familiar with what algebraic expressions
(or simply expressions) are. Examples of expressions are:
x + 3, 2y – 5, 3x
2
, 4xy + 7 etc.
In the earlier classes, we have also learnt how to add and subtract algebraic expressions.
For example, to add 7x
2
– 4x + 5 and 9x – 10, we do
7
x
2
– 4x + 5
+ 9x – 10
7x
2
+ 5x – 5
Observe how we do the addition. We write each expression to be added in a separate
row. While doing so we write like terms one below the other, and add them, as shown.
Thus 5 + (–10) = 5 –10 = –5. Similarly, – 4x + 9x = (– 4 + 9)x = 5x. Let us take some
more examples.
Example 1: Add: 7
xy + 5yz – 3zx, 4yz + 9zx – 4y , –3xz + 5x – 2xy.
Solution: Writing the three expressions in separate rows, with like terms one below
the other, we have
7xy + 5yz –3zx
+ 4yz + 9zx – 4y
+ –2xy – 3zx + 5
x (Note xz is same as zx)
5xy + 9yz + 3zx + 5x – 4y
Thus, the sum of the expressions is 5xy + 9yz + 3zx + 5x – 4y. Note how the terms, – 4y
in the second expression and 5x in the third expression, are carried over as they are,
since they have no like terms in the other expressions.
Algebraic Expressions
and Identities
CHAPTER
8
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94 MATHEMATICS
Example 2: Subtract 5x
2
– 4y
2
+ 6y – 3 from 7x
2
– 4xy + 8y
2
+ 5x – 3y.
Solution:
7x
2
– 4xy + 8y
2
+ 5x – 3y
5x
2
– 4y
2
+ 6
y – 3
(–) (+) (–) (+)
2x
2
– 4
xy + 12y
2
+ 5x – 9y + 3
Note that subtraction of a number is the same as addition of its additive inverse.
Thus subtracting –3 is the same as adding +3. Similarly, subtracting 6y is the same as
adding – 6y; subtracting – 4y
2
is the same as adding 4y
2
and so on. The signs in the
third row written below each term in the second row help us in knowing which
operation has to be performed.
EXERCISE 8.1
1. Add the following.
(i) ab – bc, bc – ca, ca – ab (ii) a – b + ab, b – c + bc, c – a + ac
(iii) 2p
2
q
2
– 3pq + 4, 5 + 7pq – 3p
2
q
2
(iv) l
2
+ m
2
, m
2
+ n
2
, n
2
+ l
2
,
2lm + 2mn + 2nl
2. (a) Subtract 4a – 7ab + 3b + 12 from 12a – 9ab + 5b – 3
(b) Subtract 3xy + 5yz – 7zx from 5xy – 2yz – 2zx + 10xyz
(c) Subtract 4p
2
q – 3pq + 5pq
2
– 8p + 7q – 10 from
18 – 3p – 11q + 5pq – 2pq
2
+ 5p
2
q
8.2 Multiplication of Algebraic Expressions:
Introduction
(i) Look at the following patterns of dots.
Pattern of dots Total number of dots
4 × 9
5 × 7
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ALGEBRAIC EXPRESSIONS AND IDENTITIES 95
m × n
(m + 2) × (n + 3)
(ii) Can you now think of similar other situations in which
two algebraic expressions have to be multiplied?
Ameena gets up. She says, “We can think of area of
a rectangle.” The area of a rectangle is l × b, where l
is the length, and b
is breadth. If the length of the
rectangle is increased by 5 units, i.e., (l + 5) and
breadth is decreased by 3 units , i.e., (b – 3) units,
the area of the new rectangle will be (l + 5) × (b – 3).
(iii) Can you think about volume? (The volume of a
rectangular box is given by the product of its length,
breadth and height).
(iv) Sarita points out that when we buy things, we have to
carry out multiplication. For example, if
price of bananas per dozen = ' p
and for the school picnic bananas needed = z dozens,
then we have to pay = ' p × z
Suppose, the price per dozen was less by ' 2 and the bananas needed were less by
4 dozens.
Then, price of bananas per dozen = ' (p – 2)
and bananas needed = (z – 4) dozens,
Therefore, we would have to pay = ' (p – 2) × (z – 4)
To find the area of a rectangle, we
have to multiply algebraic
expressions like l × b or
(l + 5) × (b – 3).
Here the number of rows
is increased by
2, i.e., m + 2 and number
of columns increased by
3, i.e., n + 3.
To find the number of
dots we have to multiply
the expression for the
number of rows by the
expression for the
number of columns.
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96 MATHEMATICS
Notice that all the three
products of monomials, 3xy,
15xy, –15xy, are also
monomials.
TRY THESE
Can you think of two more such situations, where we may need to multiply algebraic
expressions?
[Hint: • Think of speed and time;
• Think of interest to be paid, the principal and the rate of simple interest; etc.]
In all the above examples, we had to carry out multiplication of two or more quantities. If
the quantities are given by algebraic expressions, we need to find their product. This
means that we should know how to obtain this product. Let us do this systematically. To
begin with we shall look at the multiplication of two monomials.
8.3 Multiplying a Monomial by a Monomial
Expression that contains only one term is called a monomial.
8.3.1 Multiplying two monomials
We begin with
4 × x = x + x + x + x = 4x as seen earlier.
Similarly, 4 × (3x) = 3x + 3x + 3x + 3x = 12x
Now, observe the following products.
(i) x × 3y = x × 3 × y = 3 × x × y = 3xy
(ii) 5
x × 3y = 5 × x × 3 ×
y = 5 × 3 ×
x × y = 15xy
(iii) 5x × (–3y) = 5 × x × (–3) × y
= 5 × (–3) × x × y = –15xy
Note that 5 × 4 = 20
i.e., coefficient of product = coefficient of
first monomial × coefficient of second
monomial;
and x × x
2
= x
3
i.e., algebraic factor of product
= algebraic factor of first monomial
× algebraic factor of second monomial.
Some more useful examples follow.
(iv) 5x × 4x
2
= (5 × 4) × (x × x
2
)
= 20 ×
x
3
= 20x
3
(v) 5x × (– 4xyz) = (5 × – 4) × (x × xyz)
= –20 × (x × x × yz) = –20x
2
yz
Observe how we collect the powers of different variables
in the algebraic parts of the two monomials. While doing
so, we use the rules of exponents and powers.
8.3.2 Multiplying three or more monomials
Observe the following examples.
(i) 2x × 5y × 7z = (2x × 5y) × 7z = 10xy × 7z = 70xyz
(ii) 4xy × 5x
2
y
2
× 6x
3
y
3
= (4xy × 5x
2
y
2
) × 6x
3
y
3
= 20x
3
y
3
× 6x
3
y
3
= 120x
3
y
3
× x
3
y
3
= 120 (x
3
× x
3
) × (y
3
× y
3
) = 120x
6
× y
6
= 120x
6
y
6
It is clear that we first multiply the first two monomials and then multiply the resulting
monomial by the third monomial. This method can be extended to the product of any
number of monomials.
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ALGEBRAIC EXPRESSIONS AND IDENTITIES 97
TRY THESE
Find 4x × 5y × 7z
First find 4x × 5y and multiply it by 7z;
or first find 5y × 7z and multiply it by 4x.
Is the result the same? What do you observe?
Does the order in which you carry out the multiplication matter?
Example 3: Complete the table for area of a rectangle with given length and breadth.
Solution: length breadth area
3x 5y 3
x × 5y = 15xy
9y 4y
2
..............
4
ab 5bc ..............
2l
2
m 3lm
2
..............
Example 4: Find the volume of each rectangular box with given length, breadth
and height.
length breadth height
(i) 2ax 3by 5cz
(ii) m
2
n n
2
p p
2
m
(iii) 2
q 4q
2
8q
3
Solution: Volume = length × breadth × height
Hence, for (i) volume = (2ax) × (3
by) × (5cz)
= 2 × 3 × 5 × (ax) × (by) × (cz) = 30abcxyz
for (ii) volume = m
2
n × n
2
p × p
2
m
= (m
2
× m) × (n × n
2
) × (p × p
2
) = m
3
n
3
p
3
for (iii) volume = 2q
× 4q
2
× 8q
3
= 2 × 4 × 8 × q × q
2
× q
3
= 64q
6
EXERCISE 8.2
1. Find the product of the following pairs of monomials.
(i) 4, 7p (ii) – 4p, 7p (iii) – 4p, 7pq (iv) 4p
3
, – 3p
(v) 4p, 0
2. Find the areas of rectangles with the following pairs of monomials as their lengths and
breadths respectively.
(p, q); (10m, 5n); (20x
2
, 5y
2
); (4x, 3x
2
); (3mn, 4np)
We can find the product in other way also.
4xy × 5x
2
y
2
× 6x
3
y
3
= (4 × 5 × 6) × (x × x
2
× x
3
) × (y × y
2
× y
3
)
= 120 x
6
y
6
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98 MATHEMATICS
First monomial →
Second monomial ↓
TRY THESE
3. Complete the table of products.
2x –5y 3x
2
– 4xy 7x
2
y –9x
2
y
2
2x 4x
2 . . . . . . . . . . . . . . .
–5y
. . . . . .
–15x
2
y
. . . . . . . . .
3x
2 . . . . . . . . . . . . . . . . . .
– 4xy
. . . . . . . . . . . . . . . . . .
7x
2
y
. . . . . . . . . . . . . . . . . .
–9x
2
y
2 . . . . . . . . . . . . . . . . . .
4. Obtain the volume of rectangular boxes with the following length, breadth and height
respectively.
(i) 5a
, 3a
2
, 7a
4
(ii) 2p, 4q, 8r (iii) xy, 2x
2
y, 2xy
2
(iv) a, 2b, 3c
5. Obtain the product of
(i) xy, yz, zx (ii) a, – a
2
, a
3
(iii) 2, 4y, 8y
2
, 16y
3
(iv) a, 2b, 3c, 6abc (v) m, – mn, mnp
8.4 Multiplying a Monomial by a Polynomial
Expression that contains two terms is called a binomial. An expression containing three
terms is a trinomial and so on. In general, an expression containing, one or more terms with
non-zero coefficient (with variables having non negative integers as exponents) is called
a polynomial.
8.4.1 Multiplying a monomial by a binomial
Let us multiply the monomial 3x by the binomial 5y + 2, i.e., find 3x × (5y + 2) = ?
Recall that 3x and (5y + 2) represent numbers. Therefore, using the distributive law,
3x × (5y + 2) = (3x × 5y) + (3x × 2) = 15xy + 6x
We commonly use distributive law in our calculations. For example:
7 × 106 = 7 × (100 + 6)
= 7 × 100 + 7 × 6 (Here, we used distributive law)
= 700 + 42 = 742
7 × 38 = 7 × (40 – 2)
= 7 × 40 – 7 × 2 (Here, we used distributive law)
= 280 – 14 = 266
Similarly, (–3x) × (–5y + 2) = (–3x) × (–5y) + (–3x) × (2) = 15xy – 6x
and 5xy × (y
2
+ 3) = (5xy × y
2
) + (5xy × 3) = 5xy
3
+ 15xy.
What about a binomial × monomial? For example, (5y + 2) × 3x = ?
We may use commutative law as : 7 × 3 = 3 × 7; or in general a × b = b × a
Similarly, (5y + 2) × 3x = 3x × (5y + 2) = 15xy + 6x as before.
Find the product (i) 2x (3x + 5xy) (ii) a
2
(2ab – 5c)
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ALGEBRAIC EXPRESSIONS AND IDENTITIES 99
8.4.2 Multiplying a monomial by a trinomial
Consider 3p × (4p
2
+ 5p + 7). As in the earlier case, we use distributive law;
3p × (4p
2
+ 5p
+ 7) = (3p × 4p
2
) + (3p × 5p) + (3p × 7)
= 12p
3
+ 15p
2
+ 21p
Multiply each term of the trinomial by the monomial and add products.
Observe, by using the distributive law, we are able to carry out the
multiplication term by term.
Example 5: Simplify the expressions and evaluate them as directed:
(i) x (x – 3) + 2 for x = 1, (ii) 3y (2y – 7) – 3 (y – 4) – 63 for
y = –2
Solution:
(i) x (x – 3) + 2 = x
2
– 3x + 2
For x = 1, x
2
– 3x + 2 = (1)
2
– 3 (1) + 2
= 1 – 3 + 2 = 3 – 3 = 0
(ii) 3y (2y – 7) – 3 (y – 4) – 63 = 6y
2
– 21y – 3y + 12 – 63
=
6y
2
– 24y – 51
For y = –2, 6y
2
– 24y – 51 = 6 (–2)
2
– 24(–2) – 51
= 6 × 4 + 24 × 2 – 51
= 24 + 48 – 51 = 72 – 51 = 21
Example 6: Add
(i) 5m (3 – m) and 6m
2
– 13m (ii) 4y (3y
2
+ 5y – 7) and 2 (y
3
– 4y
2
+ 5)
Solution:
(i) First expression = 5m (3 – m) = (5m × 3) – (5m × m) = 15m – 5m
2
Now adding the second expression to it,15m – 5m
2
+ 6m
2
– 13m = m
2
+ 2m
(ii) The first expression = 4y (3y
2
+ 5
y – 7) = (4y × 3y
2
) + (4y × 5y) + (4y × (–7))
= 12y
3
+ 20y
2
– 28y
The second expression = 2 (y
3
– 4y
2
+ 5) = 2y
3
+ 2 × (– 4y
2
) + 2 × 5
= 2y
3
– 8y
2
+ 10
Adding the two expressions, 12y
3
+ 20y
2
– 28y
+ 2y
3
– 8y
2
+ 10
14y
3
+ 12y
2
– 28y + 10
Example 7: Subtract 3pq (p – q) from 2pq (p + q).
Solution: We have 3pq (p – q) = 3p
2
q – 3pq
2
and
2pq (p + q) = 2p
2
q + 2pq
2
Subtracting, 2p
2
q + 2pq
2
3p
2
q – 3pq
2
– +
– p
2
q + 5pq
2
TRY THESE
Find the product:
(4p
2
+ 5
p
+ 7) × 3
p
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100 MATHEMATICS
EXERCISE 8.3
1. Carry out the multiplication of the expressions in each of the following pairs.
(i) 4p, q + r (ii) ab
, a – b (iii) a + b, 7a
2
b
2
(iv) a
2
– 9, 4a
(v) pq + qr + rp, 0
2. Complete the table.
First expression Second expression Product
(i) a b + c + d
. . .
(ii) x + y – 5 5xy
. . .
(iii) p 6
p
2
– 7p + 5
. . .
(iv) 4p
2
q
2
p
2
– q
2 . . .
(v) a + b + c abc
. . .
3. Find the product.
(i) (a
2
) × (2a
22
) × (4a
26
) (ii)
2
3
9
10
2 2
xy x y
×
−
(iii)
−
×
10
3
6
5
3 3
pq p q
(iv) x × x
2
× x
3
× x
4
4. (a) Simplify 3x (4x – 5) + 3 and find its values for (i) x = 3 (ii) x =
1
2
.
(b) Simplify a (a
2
+ a + 1) + 5 and find its value for (i) a = 0, (ii)
a = 1
(iii) a = – 1.
5. (a) Add: p ( p – q), q ( q – r) and r ( r – p)
(b) Add: 2x (z – x – y) and 2y (z – y – x)
(c) Subtract: 3l (l – 4 m + 5 n) from 4l ( 10 n – 3 m + 2 l )
(d) Subtract: 3a (a + b + c ) – 2 b (a – b + c) from 4c ( – a + b + c )
8.5 Multiplying a Polynomial by a Polynomial
8.5.1 Multiplying a binomial by a binomial
Let us multiply one binomial (2a + 3b) by another binomial, say (3a + 4b). We do this
step-by-step, as we did in earlier cases, following the distributive law of multiplication,
(3a + 4b) × (2a + 3b) = 3a × (2a + 3b) + 4b × (2a + 3b)
= (3a × 2a) + (3a × 3b) + (4b × 2a) + (4b × 3b)
= 6a
2
+ 9ab + 8ba + 12
b
2
= 6a
2
+ 17ab + 12b
2
(Since ba = ab)
When we carry out term by term multiplication, we expect 2 × 2 = 4 terms to be
present. But two of these are like terms, which are combined, and hence we get 3 terms.
In multiplication of polynomials with polynomials, we should always look for like
terms, if any, and combine them.
Observe, every term in one
binomial multiplies every
term in the other binomial.
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ALGEBRAIC EXPRESSIONS AND IDENTITIES 101
Example 8: Multiply
(i) (x – 4) and (2x + 3) (ii) (x – y) and (3x + 5y)
Solution:
(i) (x – 4) × (2x + 3) = x × (2x + 3) – 4 × (2x + 3)
= (x × 2x) + (x × 3) – (4 × 2x) – (4 × 3) = 2
x
2
+ 3x – 8x – 12
= 2x
2
– 5x – 12 (Adding like terms)
(ii) (x –
y) × (3x + 5y) = x × (3x + 5y) – y × (3x + 5y)
= (x × 3x) + (x × 5y) – (y × 3x) – ( y × 5y)
= 3x
2
+ 5xy – 3yx – 5y
2
= 3x
2
+ 2xy – 5y
2
(Adding like terms)
Example 9: Multiply
(i) (a + 7) and (b – 5) (ii) (a
2
+ 2b
2
) and (5a – 3b)
Solution:
(i) (a + 7) × (b – 5) = a × (b – 5) + 7 × (b – 5)
= ab – 5a + 7b – 35
Note that there are no like terms involved in this multiplication.
(ii) (a
2
+ 2b
2
) × (5a – 3b) = a
2
(5a – 3b) + 2b
2
× (5a – 3b)
= 5a
3
– 3a
2
b + 10ab
2
– 6b
3
8.5.2 Multiplying a binomial by a trinomial
In this multiplication, we shall have to multiply each of the three terms in the trinomial by
each of the two terms in the binomial. We shall get in all 3 × 2 = 6 terms, which may
reduce to 5 or less, if the term by term multiplication results in like terms. Consider
( 7)
binomial
a +
×
2
( 3 5)
trinomial
a a
+ +
= a × (a
2
+ 3a + 5) + 7 × (a
2
+ 3
a + 5)
= a
3
+ 3a
2
+ 5a + 7a
2
+ 21a + 35
= a
3
+ (3a
2
+ 7a
2
) + (5a + 21a) + 35
= a
3
+ 10a
2
+ 26a + 35 (Why are there only 4
terms in the final result?)
Example 10: Simplify (a + b) (2a – 3b + c) – (2a – 3b) c.
Solution: We have
(a + b) (2a – 3b + c) = a (2a – 3b + c) + b (2a – 3b + c)
= 2a
2
– 3ab + ac + 2ab – 3b
2
+ bc
= 2
a
2
– ab – 3b
2
+ bc + ac (Note, –3ab and 2ab
are like terms)
and (2a – 3b) c = 2ac – 3bc
Therefore,
(a + b) (2a – 3b + c) – (2a – 3b) c = 2a
2
– ab – 3b
2
+ bc + ac – (2ac – 3bc)
= 2a
2
– ab – 3b
2
+ bc + ac – 2ac + 3bc
= 2a
2
– ab – 3b
2
+ (bc + 3bc) + (ac – 2ac)
= 2a
2
– 3b
2
– ab + 4bc – ac
[using the distributive law]
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102 MATHEMATICS
EXERCISE 8.4
1. Multiply the binomials.
(i) (2x + 5) and (4x – 3) (ii) (y – 8) and (3y – 4)
(iii) (2.5l – 0.5m) and (2.5l + 0.5m) (iv) (a
+ 3b) and (x + 5)
(v) (2pq + 3q
2
) and (3pq – 2q
2
)
(vi)
2. Find the product.
(i) (5 – 2
x) (3
+ x) (ii) (x + 7y) (7
x – y)
(iii) (a
2
+ b
) (a + b
2
) (iv) (p
2
– q
2
) (2p + q)
3. Simplify.
(i) (x
2
– 5) (x + 5) + 25 (ii) (a
2
+ 5) (b
3
+ 3) + 5
(iii) (t + s
2
) (t
2
– s)
(iv) (a + b) (c – d) + (a – b) (c + d) + 2 (ac + bd)
(v) (x + y)(2x + y) + (x + 2y)(x – y) (vi) (x + y)(x
2
– xy + y
2
)
(vii) (1.5x – 4y)(1.5x + 4y + 3) – 4.5x + 12y
(viii) (a + b + c)(a + b – c)
WHAT HAVE WE DISCUSSED?
1. Expressions are formed from variables and constants.
2. Terms are added to form expressions. Terms themselves are formed as product of factors.
3. Expressions that contain exactly one, two and three terms are called monomials, binomials and
trinomials respectively. In general, any expression containing one or more terms with non-zero
coefficients (and with variables having non- negative integers as exponents) is called a polynomial.
4. Like terms are formed from the same variables and the powers of these variables are the same,
too. Coefficients of like terms need not be the same.
5. While adding (or subtracting) polynomials, first look for like terms and add (or subtract) them;
then handle the unlike terms.
6. There are number of situations in which we need to multiply algebraic expressions: for example, in
finding area of a rectangle, the sides of which are given as expressions.
7. A monomial multiplied by a monomial always gives a monomial.
8. While multiplying a polynomial by a monomial, we multiply every term in the polynomial by the
monomial.
9. In carrying out the multiplication of a polynomial by a binomial (or trinomial), we multiply term by
term, i.e., every term of the polynomial is multiplied by every term in the binomial (or trinomial).
Note that in such multiplication, we may get terms in the product which are like and have to be
combined.
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